$f(z)$ is holomorphic, prove that $g(z) = \overline{f(\overline{z})}$ is holomorphic.

Question

$f(z)$ is holomorphic, prove that $g(z) = \overline{f(\overline{z})}$ is holomorphic.

My attempt

With $f(z) = f(x+iy) = u(x,y) + iv(x,y)$ we get

$$g(z) =\overline{f(\overline{z})} = \overline{f(x+i(-y))} = u(x,-y) - iv(x,-y).$$

Since $f$ is holomorphic, it's true that $$u'_x(x,y) = v'_y(x,y),$$ $$u'_y(x,y) = -v'_y(x,y).$$ For $g$ to be holomorphic the following must be true $$u'_x(x,-y) = -v'_y(x,-y)$$ $$u'_y(x,-y) = v'_x(x,-y)$$

Using the Cauchy-Riemann equations for $f$ I get that $$v'_y(x,-y) = -v'_y(x,-y),$$ $$-v'_x(x,-y) = v'_x(x,-y),$$ must be true for $g$ to be holomorphic, but this can only true for $v'_x = v'_y = 0$ no?

How can I make my solution work? Is there a better (simple) way?

Answer 1

Your method should work, if you took the partial derivatives correctly. You should use the chain rule for partial derivatives, there lies your mistake.

Your function is $g(x,y) = u(x,-y)+ i( - v(x,-y))= a(x,y) + i b(x,y)$. (We use other letters so we are not confused). Now calculate the partials:

$$ \frac{\partial a}{\partial x}(x,y) = \frac{\partial }{\partial x}( u(x,-y)) = u_x(x,-y) \\ \frac{\partial a}{\partial y}(x,y) = \frac{\partial }{\partial y}( u(x,-y)) = -u_y(x,-y)\\ \frac{\partial b}{\partial x}(x,y) = \frac{\partial }{\partial x}( -v(x,-y)) =-v_x(x,-y)\\ \frac{\partial b}{\partial y}(x,y) = \frac{\partial }{\partial y}( -v(x,-y)) = v_y(x,-y) $$

It should be clear now how to use the C-R equations.

Answer 2

You can prove it by definition:

$\dfrac{g(z+h)-g(z)}{h}=\dfrac{\overline{f}(\overline{z}+\overline{h})-\overline{f}(\overline{z})}{h}=\overline{\Big(\dfrac{f(\overline{z}+\overline{h})-f(\overline{z})}{\overline{h}}\Big)}$. Using that $f$ is differentiable at $\overline{z}$, you are done.

Answer 3

Note that $f(z)$ is holomorphic iff $\frac{df(z)}{d\bar{z}}=0$ or $\frac{df(\bar{z})}{dz}=0$. So $$ \frac{dg(z)}{d\bar{z}}=\frac{d\overline{f(\bar{z})}}{d\bar{z}}=\overline{\left(\frac{df(\bar{z})}{dz}\right)}=0 $$ and hence $g(z)$ is holomorphic.